Recent content by Esoremada

Homework Statement I am trying to graph the flux density field between two infinite line charges located at y = 1 and y = -1 Homework Equations I am trying to do it using the equation for a line charge that I got from lecture notes. The above equation is derived from this: Here is the...

Oh man, messed up that equation after using it hundreds of times. Thank you very much, that fixed it. (and I forget the 1/2 both times and the mistakes canceled out which is why I got the right answer for part b :rolleyes:)

No I used horizontal velocity and distance to find time, and then used vertical velocity, distance and time to find acceleration. I'll add subscripts to the OP to make that more clear I took the electron values to many more decimals than the homework system checks (3). I changed all my...

Homework Statement http://puu.sh/6i6HY.png Homework Equations Fe = q*Ee The Attempt at a Solution a) dx = vx*t 0.0682 = 4.92*106 * cos(68) * t t = 0.0682 / [4.92*106 * cos(68)] = 3.70036*10-8 seconds dy = dyi + vy*t + at2 0 = 0 + 4.92*106 * [0.0682 / [4.92*106 * cos(68)]] * sin(68) + a...

Alright thanks! I've been thinking of torque too much in terms of physical movement rather than the vectors themselves, I think I get it now.

Right, I understand that solution, but I don't understand what the problem is with my reasoning that gives the answer into the page: The entire system has no torque to begin with, just angular momentum The axle itself is then tilted to the right (from the point of view of the diagram) This tilt...

Homework Statement http://puu.sh/5Qrxt.jpg Homework Equations τ = dL/dt The Attempt at a Solution I understand the initial and final angular momentum vectors in the diagram, and the answer seems to make sense. But when I initially solved it my logic was that since the axle...

Homework Statement http://puu.sh/5rfl9.png Homework Equations CM = Σmr / Σm L = Σ (r x p) The Attempt at a Solution CM = 0.118 * (0.0036 + 0.00657) / (0.0723 + 0.118) = 0.00630614818 m (from the centre of top disk) L = Σ (r x p) = [0.118 * (0.0036 + 0.00657) /...

...I need more sleep. Thanks again :P

Oh cool I never even noticed that. I get how to derive 1/3*ML^2 now but I try doing it in this problem and get it wrong 1250*(67.102 - 99/2)^2 + 1/12*1250*99^2 That's the moment of the rod rotating about its centre plus the moment of the rod's centre of mass rotating about the systems centre...

Oh wow I left out the 1/2 on the disks that was dumb. I fixed that and got the right answer. My initial post was treating the annulus as a difference of two disks. How could I do this without finding the mass density? I need the mass of the two disks.

That was correct! So what is the correct way to calculate the moment of a inertia of a rod about a given pivot? I only know that it is 1/12*ML^2 and 1/3*ML^2 for pivots about the centre or end.

The CoM moves through space, but how can that affect this question when the distance from the centre of mass to the ships remains the same regardless of orientation and speed relative to space.

Homework Statement http://puu.sh/5nLnI.png Homework Equations I = mr2 The Attempt at a Solution Moment of inertia: Mass density = mass / area = 2.59 / (π*9.52 - π*7.52) = 0.02424772368 kg / cm2I = (areaA)(density)(radiusA) - (areaB)(density)(radiusB) =...

Homework Statement http://puu.sh/5nKxl.png Homework Equations α = ω/t τ = I*α The Attempt at a Solution CM = [232200*99 + 99/2*12500] / (107100 + 232200 + 12500) = 67.102 m τ = 43320*67.102 + 43320*(99 - 67.102) = 4288680 N α = ω/t τ = I*α τ = I*(ω/t) ω...